Question 949773
{{{(4y-3) /(15x^2-30)=n/(30x^2-60)}}}... if you look closely at denominators you see that coefficient and constant of {{{(30x^2-60)}}} is two times greater then {{{(15x^2-30)}}}; so, to have left side of equation same as right side, all you need is two multiply numerator {{{(4y-3)}}} by {{{2}}} and you will get a numerator {{{n}}} which will be {{{2(4y-3)=8y-6}}}-equivalent expression 

so, the rational expression is:

{{{(4y-3) /(15x^2-30)=(8y-6)/(30x^2-60)}}}



or, you can solve it using proportion rules:


{{{(4y-3) /(15x^2-30)=n/(30x^2-60)}}}


{{{n=((4y-3) /(15x^2-30))(30x^2-60)}}}


{{{n=(4y-3) /(15(x^2-2))30(x^2-2)}}}


{{{n=(cross(30)2(4y-3)cross((x^2-2))) /(cross(15)cross((x^2-2)))}}}


{{{n=2(4y-3) /1}}}


{{{n=(8y-6)}}} 

answer is:

{{{(4y-3) /(15x^2-30)=(8y-6) /(30x^2-60)}}}